Realisability cohomological class as product or as immersed sphere - MathOverflow

Realisability cohomological class as product or as immersed sphere

2011-10-22T10:45:07Z

Let's consider closed simply connected manifold $M^n$ and a $a\in H^k(M)$ and $a*\in H^{n-k}(M)$ is the dual to $a$.

Is it true that dual to $a$ is realisable as a immersed sphere or $ a*=bc $ for some $b,c\in H^*(M)$ ?

Edit: it is more natural to ask about possibility to decompose dual to $a$ as a product, see example in the answer below.

Edit2: let's assume that $M$ is not decompasable, so there is no $X,Y$ such that $M = X\times Y$

Answer by Sergey Melikhov for Realisability cohomological class as product or as immersed sphere

2011-10-22T16:18:20Z

I will construct a closed simply-connected $8$-manifold $M$ and an $a\in H^3(M;\Bbb Z)$ such that the Poincare dual $b$ of $a$ is not realizable by a map $S^5\to M$, and a Hom-dual element in $H^5(M;\Bbb Z/2)$ to the $\bmod 2$ reduction of $b$ is not a nontrivial product.

Let $K$ be the suspension over $\Bbb C P^2$. Then there is an $\alpha\in H^3(K;\Bbb Z/2)$ with $Sq^2(\alpha)\ne 0$, but the $\bmod 2$ cohomology ring of $K$ is trivial. Let $N$ be a regular neighborhood of a PL copy of $K$ in some $\Bbb R^m$. So $N$ is homotopy equivalent to $K$. A loop in $\partial N$ bounds a disk in $N$, which can be pushed off $K$ as long as $5+2\le m-1$. Thus $\partial N$ is simply-connected. Let $M$ be the double of $N$, i.e. $M=\partial (N\times I)$. So $M$ is a closed $m$-manifold, it is simply-connected by Seifert-van Kampen, and the inclusion $N\subset M$ is split by the projection $\phi:M\subset N\times I\to N$. So the cohomology of $N\simeq K$ is a direct summand in the cohomology of $M$. Let $\beta=\phi^\ast(\alpha)$, then $\gamma:=Sq^2\beta\ne 0$, and $\gamma$ is not a nontrivial product. Since the nonzero element of $H^5(S^5;\Bbb Z/2)$ is not in the image of $Sq^2$, there is no map $f:S^5\to M$ such that $f^*(\gamma)\ne 0$.

Let $b\in H_5(M;\Bbb Z)$ be such that $\gamma(b)\ne 0$, and let $a\in H^{m-5}(M;\Bbb Z)$ be the Poincare dual of $b$. If $b$ is realized by an immersion, or just a map, $f:S^5\to M$, then $0\ne\gamma\smallfrown f_\ast[S^5]=f_\ast(f^\ast(\gamma)\smallfrown[S^5])$, contradicting $f^*(\gamma)=0$.

As for $m$, $\Bbb C P^2$ is the mapping cone of the Hopf map $h:S^3\to S^2$. The mapping cylinder of $h$ embeds in $S^2*S^3=S^6$, so $\Bbb CP^2$ embeds in $\Bbb R^7$ and $K$ in $\Bbb R^8$.

Answer by Somnath Basu for Realisability cohomological class as product or as immersed sphere

2011-10-23T01:13:43Z

Consider $M=SU(4)$. Rationally, the cohomology ring $H^\ast(M;\mathbb{Q})$ is the product $H^\ast(S^3;\mathbb{Q})\otimes H^\ast(S^5;\mathbb{Q})\otimes H^\ast(S^7;\mathbb{Q})$. Consider $a$ to the product of the generators of $S^3$ and $S^5$. The dual is the generator of $S^7$ which doesn't split. However, $\pi_8(SU(4))=\mathbb{Z}_{4!}$ which is torsion and hence a sphere cannot possibly generate $a$.